(a) What does the question mean by 'determining the kernel'? I know that kernels are ideals and ideals are kernels. Since $\displaystyle \varphi$ is a ring homomorphism from $\displaystyle R \to \mathbb{Z}$, then $\displaystyle ker(\varphi)=\{ r \in R | \varphi (r) = 0 \}$ is an ideal of R.

Now, a subring I is an ideal of R if $\displaystyle rI := \{ ra: a \in I \} \subseteq I$ and $\displaystyle Ir := \{ ar : a \in I \} \subseteq I$ for $\displaystyle \forall r \in R$. (The subring I is closed under multiplication with arbitrary elements of R). Is this all I need to say?!

(b) I have no idea how to do this one...

(c) I think $\displaystyle ker(\varphi)$ is a prime ideal of R, if it is a proper ideal of R such that $\displaystyle a,b \in R$ and $\displaystyle ab \in ker(\varphi)$ imply $\displaystyle a \in ker(\varphi)$ or $\displaystyle b \in ker(\varphi)$. What else do I need to state?

(a) What does the question mean by 'determining the kernel'? I know that kernels are ideals and ideals are kernels. Since $\displaystyle \varphi$ is a ring homomorphism from $\displaystyle R \to \mathbb{Z}$, then $\displaystyle ker(\varphi)=\{ r \in R | \varphi (r) = 0 \}$ is an ideal of R.

Now, a subring I is an ideal of R if $\displaystyle rI := \{ ra: a \in I \} \subseteq I$ and $\displaystyle Ir := \{ ar : a \in I \} \subseteq I$ for $\displaystyle \forall r \in R$. (The subring I is closed under multiplication with arbitrary elements of R). Is this all I need to say?!

No. I mean, you've just given the definition of a kernel. You need to apply it to this case. It's just the set of matrices such that a-3b=0. So just stick that in set notation...

Quote:

Originally Posted by demode

(b) I have no idea how to do this one...

Okay, you have a ring homomorphism which maps into the integers. Basically, you are trying to show that the image is isomorphic to the integers (this is easiest, and equivalent). So, how would you approach this problem? You know that the image injects into the integers (because it is a subring). So, what is left to show?...

Quote:

Originally Posted by demode

(c) I think $\displaystyle ker(\varphi)$ is a prime ideal of R, if it is a proper ideal of R such that $\displaystyle a,b \in R$ and $\displaystyle ab \in ker(\varphi)$ imply $\displaystyle a \in ker(\varphi)$ or $\displaystyle b \in ker(\varphi)$. What else do I need to state?

This is a exercise in looking at the image to find things about the pre-image/kernel. So, if $\displaystyle ker(\varphi)$ is a prime ideal, then what does this mean about the integers? If $\displaystyle ker(\varphi)$ is a maximal ideal what does this mean about the integers?

Sep 21st 2010, 05:08 PM

demode

Quote:

Originally Posted by Swlabr

No. I mean, you've just given the definition of a kernel. You need to apply it to this case. It's just the set of matrices such that a-3b=0. So just stick that in set notation...

$\displaystyle ker(\varphi)= \{ a-3b=0 | a,b \in \mathbb{Z} \}$

Quote:

Okay, you have a ring homomorphism which maps into the integers. Basically, you are trying to show that the image is isomorphic to the integers (this is easiest, and equivalent). So, how would you approach this problem? You know that the image injects into the integers (because it is a subring). So, what is left to show?...

So, I just need to show that $\displaystyle R/ker(\varphi)$ is surjective and operation preserving?

Quote:

This is a exercise in looking at the image to find things about the pre-image/kernel. So, if $\displaystyle ker(\varphi)$ is a prime ideal, then what does this mean about the integers? If $\displaystyle ker(\varphi)$ is a maximal ideal what does this mean about the integers?

What do you mean by that? Could you please explain a little bit more?

Sep 21st 2010, 11:14 PM

Swlabr

Quote:

Originally Posted by demode

So, I just need to show that $\displaystyle R/ker(\varphi)$ is surjective and operation preserving?

You have operation preserving - that's just the first Isomorphism theorem. So you just need to prove surjective. Can you think of a neat way of proving that the homomorphism is surjective?

I tried to follow your hint, but I still need some help to wrap it up:

If $\displaystyle I$ is a prime ideal in $\displaystyle R$ then: $\displaystyle R/I$ is an integral domain if and only if $\displaystyle I$ is a prime ideal. (Also $\displaystyle R/I$ is a field if and only if $\displaystyle I$ is maximal.)

$\displaystyle R/I$ is clearly commutative and it has identity $\displaystyle 1+I$. In order to show it's an integral domain I only need to show that there are no zero divisors.

x+I and y+I are nonzero in $\displaystyle R/I$ $\displaystyle \iff$ $\displaystyle x$ and $\displaystyle y$ do not lie in $\displaystyle I$. If I is a prime then x.y dos not lie in I and so $\displaystyle (x+I)(y+I)=x.y+I$ must be non-zero.

Use the fact that $\displaystyle R/I \cong \mathbb{Z}$. No working is really needed then.

Quote:

Originally Posted by demode

I see, but I really can't think of a neat way of doing it... maybe some more hints please?

You need to prove that the generators (or here, generator) is mapped to. Can you see why this is sufficient?

Sep 24th 2010, 04:28 AM

demode

Quote:

Originally Posted by Swlabr

Use the fact that $\displaystyle R/I \cong \mathbb{Z}$. No working is really needed then.

I'm not sure how to fit this in, could you please show me where it can be used? Here's the proof I've written so far:

Suppose I is a prime and suppose R/I is not an integral domain. There are zero divisors in R/I iff there are $\displaystyle (x+I), (y+I) \in (R/I)^*$. So that $\displaystyle (x+I)(y+I) = 0_R =I \in (R/I)^*$.

So, $\displaystyle x \notin I $and $\displaystyle y \notin I$ but $\displaystyle xy \in I$. So we get a contradiction and R/I must be an integral domain and so $\displaystyle I=ker(\varphi)$ is a prime.

Quote:

You need to prove that the generators (or here, generator) is mapped to. Can you see why this is sufficient?

I'm not sure how to fit this in, could you please show me where it can be used? Here's the proof I've written so far:

Suppose I is a prime and suppose R/I is not an integral domain. There are zero divisors in R/I iff there are $\displaystyle (x+I), (y+I) \in (R/I)^*$. So that $\displaystyle (x+I)(y+I) = 0_R =I \in (R/I)^*$.

So, $\displaystyle x \notin I $and $\displaystyle y \notin I$ but $\displaystyle xy \in I$. So we get a contradiction and R/I must be an integral domain and so $\displaystyle I=ker(\varphi)$ is a prime.

Because the two rings are isomorphic, if one has zero divisors then so does the other. Similarly, if one is a field then so is the other. Apply these two facts...

You want to show that 1 is contained in $\displaystyle im(\varphi)$. That is, does there exist some 2-by-2 matrix of that form which has determinant equal to 1?

Oct 30th 2010, 08:50 PM

demode

Since the factor ring $\displaystyle R/ ker(\varphi)$ is an integral domain, it follows kernel is a prime ideal. But how do you show that $\displaystyle ker(\varphi)$ is not maximal? More precisely, how do you show that $\displaystyle R/ ker(\varphi)$ is not a field?

Nov 1st 2010, 12:53 AM

Swlabr

Quote:

Originally Posted by demode

Since the factor ring $\displaystyle R/ ker(\varphi)$ is an integral domain, it follows kernel is a prime ideal. But how do you show that $\displaystyle ker(\varphi)$ is not maximal? More precisely, how do you show that $\displaystyle R/ ker(\varphi)$ is not a field?

You are given that $\displaystyle R/ ker(\varphi) \cong \mathbb{Z}$...